+0  
 
0
50
3
avatar

Please provide indepth explination thanks!

Let $\triangle ABC$ be a right triangle, with the point $H$ the foot of the altitude from $C$ to side $\overline{AB}$.

Prove that 

Guest Oct 29, 2018
 #1
avatar+3270 
+1

Try to expand the terms! 

tertre  Oct 30, 2018
 #2
avatar+963 
+3

Expanding the terms, 

 

\((x^2+2xh+h^2)+(y^2+2yh+y^2)=(a^2+2ab+b^2) \)

 

Using Pythagorean Theorem,

 

\(x^2+h^2=a^2\\ y^2+h^2=b^2\)

 

We could subsitute the values in, and rewrite the equation

 

\(a^2+2xh+b^2+2yh=a^2+2ab+b^2\\ 2xh+2yh=2ab\\ h(x+y)=ab \)

 

\([ABC]=\frac12 AB\cdot CH = \frac12 BC \cdot AC\\ \frac12 (x+y)h=\frac12 ab\\ h(x+y)=ab\)

 

I hope this helped,

 

Gavin

GYanggg  Oct 30, 2018
edited by GYanggg  Oct 30, 2018
 #3
avatar+91027 
+2

Note that triangle  BCA  is similar to triangle  BHC

 

Which implies that

 

HC / BC  =  CA / BA..... so...

 

h / a  =  b / ( x + y)    (1)

 

Now expand   ( x + h)^2 + ( y + h)^2  =  ( a + b)^2    (2)

 

x^2 + 2xh + h^2  + y^2 + 2yh + h^2  =  a^2 + 2ab + b^2     (3) 

 

And since  x^2 + h^2  =  a^2    and   y^2 + h^2  = b^2

 

We can subtract these equal parts from  (3)  and we are left with

 

2xh + 2yh  =  2ab        divide through by 2

 

xh + yh  =  ab      factor out h on the left

 

h ( x + y)  =  ab       rearrange as

 

h / a  =  b / ( x + y)     but, by (1)....this is true

 

So (2)  must be true, as well

 

cool cool cool

CPhill  Oct 31, 2018

27 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.